Question: Subtract. $\dfrac{8}{3} - \dfrac{2}{8} = $
Explanation: Before we can subtract our fractions, they need to have the same denominator. $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{3}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\dfrac{8}{3}$ $\dfrac{2}{8}$ $\dfrac{8}{3}-\dfrac{2}{8}$ Let's look at the multiples of each denominator and see which multiples they have in common. Denominator Multiples ${3}$ $3, {6}, 9, 12, 15, 18, 21, \underline{24}$ $8}$ $8, 16, \underline{24}, 32$ The least common denominator is ${24}$. Let's use multiplication to make each fraction have a denominator of $24$. ${\dfrac{8}{3}}=\dfrac{{8} \times 8}{{3} \times 8} = {\dfrac{64}{24}}$ $\dfrac{2}{8}}=\dfrac{2} \times 3}{8} \times 3} = {\dfrac6}24}}$ Now, we can subtract ${\dfrac{64}{24}} - \dfrac{6}{24}}$. $\dfrac{64}{24}$ $\dfrac{6}{24}$ $\dfrac{64}{24} - \dfrac{6}{24}$ $=\dfrac{{64}-6}}{24}$ $= \dfrac{58}{24}$ ${\dfrac{8}{3}} - \dfrac{2}{8}} = \dfrac{58}{24}$ We can also write $\dfrac{58}{24}$ as $\dfrac{29}{12}$ or $2\dfrac{5}{12}$.